Can anyone see how the solution to the differential equation $$\frac{dc(t)}{dt} = -c(t)\bigg\langle \psi(t) \left \lvert \frac{\partial \psi(t)}{\partial t } \bigg\rangle \right .$$ has the solution $c(t) = c(0)e^{i \gamma(t)}$ where $$\gamma(t) := i \int_{0}^{t} \bigg\langle \psi(t') \left \lvert \frac{\partial}{\partial t'} \psi(t') \bigg\rangle dt' \right. .$$
I get the solution $c(t) = c(0)e^{-\left \langle \psi \, \left \lvert \frac{\partial \psi}{\partial t} \right \rangle t \right .}$. Where $\langle \cdot, \cdot \rangle$ is the inner product of the functions.
Thanks for any assistance.
Define
$$ f(t) = i\langle \psi(t) | \partial_t \psi(t) \rangle $$ Such that
$$ \frac{{\rm d}c(t)}{{\rm d}t} = if(t) c(t) $$
which can be rearranged as
$$ \frac{{\rm d}c(t)}{c(t)} = if(t){\rm d}t $$
And integrating at both sides
$$ \ln \frac{c(t)}{c(0)} = i\int_0^t {\rm d}t'f(t') = i\gamma(t) ~~~\Rightarrow~~~ c(t) = c(0)e^{i\gamma(t)} $$